Part 1
Calculate confidence intervals at the 95% level using the Penn State football championship voting 80/20 split examples on Page 2 and use samples of 50, 500, 2000, and 5000 to see the point of diminishing returns for yourself.
Part 2
a. P = 75% , Q = 25% n= 678 @ 95% level of confidence
b. P = 75%, Q = 25% n= 35 @ 95% level of confidence
c. P = 52%, Q = 48% n= 678 @ 99% level of confidence
A dog breeder specializes in golden retrievers. The local police department suspects that this breeder is selling puppies with an abnormal amount of hip problems to out of state customers. The department learns that the probability of a golden retriever pup developing hip problems in its first year of life is 10% (.10). How many pups out of a litter of 5 must develop hip problems for the police department to take action against this breeder with 95% confidence that she is guilty?
Assume that we are estimating 2 different basketball players' free throw shooting accuracy. The first basketball player is on a collegiate team and plays the sport almost daily year round. After watching that player shoot, we estimate that she will make 85% (.85) of her free throws. The second player, a doofus college professor with little athletic ability, is similarly observed and his free throw average is estimated at 15%. How many free throws must the first player make in a row before you can confidently reject the following null hypothesis:
Ha: The player's free throw percentage was incorrectly estimated
Ho: The player's free throw percentage was not incorrectly estimatedHow many free throws must the second player make in a row in order to confidently reject the same null hypothesis? Use the 95% level of confidence for both players.
Submit your answers to Lesson 6 More Hypothesis Testing drop box.
Part 3
We are looking at the question to examine the relationship between different news networks and their reporting practices.
Question: Are news networks actually biased in the reporting in favor of either the Republican or Democratic Party?
News Network 
# of negative stories about Democrats in first 6 months of 2006

# of negative stories about Republicans in first 6 months of 2006

Totals

Fox News

266

255

521

MSNBC

302

325

627

CNN

283

298

581

ABC

211

230

441

CBS

275

291

566

Totals

1337

1399

2736

News Network 
# of negative stories about Democrats in first 6 months of 2006

# of negative stories about Republicans in first 6 months of 2006

Totals

Fox News

266

255

521

MSNBC

302

325

627

CNN

283

298

581

ABC

211

230

441

CBS

275

291

566

Totals

1337

1399

2736

H_{a}: The major news networks favor one political party over the other in their reporting practices.
H_{o}: The major news networks do not favor one political party over the other in their reporting practices.
The ZScore as a Test of Significance
With above in mind, we may turn to testing the significance of our 8% point difference between drug tested pretrial releasees and nondrug tested releasees. Given that this issue has not been previously tested (in the study municipality), we will structure our hypotheses as nondirectional and indicative of a twotailed test of significance. The hypotheses are:
Ha: The percentages of offenders who undergo pretrial drug screening fail to appear at scheduled court proceedings at differing rates than do offenders who were not subject to drug testing prior to their court dates
Ho: The percentages of offenders who undergo pretrial drug screening fail to appear at scheduled court proceedings do not differ significantly from the rates of offenders who were not subject to drug testing prior to their court dates
In order to perform this test we need a bit more information, namely the numbers of subjects in each of the two population groups:
Drug  Tested NonDrug Tested% failing to appear 38%, or .38 30% or .30 Number of Subjects (N) 116 118 With the above information of percentage (converted to proportions to make calculation of z easier) failing to appear and population size, we can calculate the denominator of our zformula. When using the zformula as a test of significance between two groups, there is no 'standard deviation' per se (because we are using nominal data and have two separate groupings), therefore we calculate a term of measurement error which substitutes as the standard deviation we used in the formula for z to compare a single value to the population as a whole (recall the z formula below).
X= individual score
= population mean
S = population standard deviationCalculation of this denominator is a fourstep process using the percentages (converted to proportions) and the population figures.
Calculate the Denominator
Obtain a weighted proportion P.
The first step is to obtain a weighted proportion, based on the proportions from both groups as well as their respective populations:
P = (N1 X P1) + (N2 X P2) / N1 + N2
P = (118 X 0.30) + (116 X 0.38) / 116 + 118
P = 35.4 + 44.08 / 234
P = 79.48 / 234
P = 0.3397, ROUNDED TO 0.34
The Decision
We now have the denominator of the z formula so we can calculate a zscore:
We have a calculated z score of 1.29, meaning that the two percentages are 1.29 standard units apart. You'll remember that we are conducting a twotailed test of significance, given our nondirectional hypothesis. Practically, this implies that the rejection region be split between the left and right sides of the distribution mean (i.e. includes both "tails"). At the 95% level of confidence, only 5% of the area under the normal curve contains values that would lead us to reject our stated null hypothesis and conditionally accept our research hypothesis. In order to make our final decision, we compared our calculated value of z with a critical value of z obtained from the z table (in much the same fashion as we did for chisquare). The critical value is a function of both significance level (95% in this example) and directionality of the research hypothesis. At the 95% level of confidence, there is 5% 'rejection region' under the normal curve. This 5% is divided between the twotails of the normal distribution (2.5% on either side). In proportional terms .4750 of the area on either side of the mean of the normal distribution is 'nonrejection region' (.5  .025), while .025 is 'rejection region.' The critical value is that value of z occurring exactly where the rejection region starts under the normal curve.
We have an area 'cutoff' of .4750, therefore we can locate this area in the z table and find the corresponding zvalue. Usingcolumn b ('Area between X and Z'), we see that an area of .4750 corresponds with a z score of 1.96.
Comparing this critical value of z with the value of 1.29 we calculated using the z formula, we decide to fail to reject the null hypothesis (calculated value of z ‹ critical value of z), and conclude that drug testing pretrial defendants is not significantly associated with rates of failing to appear in court.
 It is thought that one tangible benefit of increasing the educational requirements for incoming police officers would be a decrease in the number of police brutality complaints filed by citizens each year. To test this idea, we can use Z to test two independent samples of police officers  one group with college degrees, the other without. A random sample of officers was selected from one state.
Please submit your solution (stepbystep calculations ), and the conclusion, at the 95% level of confidence, to theLesson 9 z Scores drop box. The data and hypotheses are as follows:
% of College educated officers who had an abuse complaint filed against them in the previous 12 months % of Noncollege educated officers who had an abuse complaint filed against them in the previous 12 months 15 18 N=485 N=612H_{a}: Collegeeducated officers will have fewer abuse complaints filed against them
H_{o}: Their levels of education do not affect the proportion of abuse complaints filed against police officers
 Much has been made about the overall intelligence of criminal offenders in general, and violent offenders in particular. There is an assumption that violent offenders are inherently less intelligent than nonviolent offenders. To test this idea, data were collected from two separate groups of offenders (violent offenders and property offenders). Intelligence scores were obtained for every offender in the two samples.
Please submit your solution (stepbystep calculations ), and the conclusion to the Lesson 9 t Test drop box. Descriptive Statistics appear below:
Property Offenders  Average IQ = 96, sd = 8.5, n=225
Violent Offenders  Average IQ = 88, sd = 7.5, n=200
Use t to test the following set of hypotheses at the 95% level of confidence:
H_{a}: Violent offenders are less intelligent than property offenders
H_{o}: Violent offenders and property offenders do not differ significantly with regard to intelligence scores
 Below are data from selected inmates classified into one of three security levels at one correctional facility, relative to their depression scores on a standard measurement instrument.
 Using the F statistic , determine if depression scores differ significantly between the three security levels at the prison.
 Plase submit your solution (stepbystep) and the conclusion (at the 95% level of confidence) to the Lesson 10 F Testdrop box.
H_{a}: Mean Depression Scores of Inmates at Different Security Levels are Significantly Different.
H_{o}: Mean Depression Scores of Inmates at Different Security Levels are NOT Significantly Different.
Low security (n=4) Medium security (n=4) High security (n=4) 3 9 9 5 9 10 4 8 7 4 6 10  16 32 36
Subject  Science 
Due By (Pacific Time)  04/15/2014 12:00 am 
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